# Is $E=\hbar \omega$ correct for massive particles?

From Planck's relation we can say that the energy of a photon is

$$E=h\nu=\hbar \omega \, .$$

where $$\hbar \equiv h / 2\pi$$. On the other hand, the energy of a free particle can be expressed as

$$E=\frac{p^2}{2m}$$

and from de Broglie relation we have

$$p = \frac{h}{\lambda} = \hbar k \, .$$

So, we can write the dispersion relation

$$\omega = \frac{\hbar k^2}{2m} \, .$$

Is this correct? We are mixing a photon energy with a particle energy. The energy of a particle in its most general way is:

$$E = \sqrt{p^2 c^2 + m^2c^4} \, .$$

For a photon we have $$m=0$$, which implies $$E = pc = h\nu$$, but this doesn't happen for a massive particle. So, is there something that I'm missing?

• Commented Feb 23, 2021 at 2:41

$\renewcommand{\ket}[1]{|#1\rangle}$ It is correct that the kinetic energy of a massive particle in the non-relativistic limit is $$E = p^2 / 2m \, . \tag{1}$$ It is also correct that for plane waves (i.e. free particle eigenstates), the momentum is related to the wave number via $$p = \hbar k \, . \tag{2}$$ Therefore, as proposed in the question, the frequency of a free massive particle in a plane wave state and in the non-relativistic limit, is $$\omega = \frac{\hbar k^2}{2m} \, . \tag{3}$$
Actually yes, it is correct. The relation $$E = h \nu = \hbar \omega \tag{4}$$ is actually a very general relation in non-relativistic quantum mechanics, not limited to photons. From one point of view, this relation comes from Schrodinger's equation for the time evolution of a quantum state $$i \hbar \frac{d \ket{\Psi}}{dt} = H \ket{\Psi} \, . \tag{5}$$ If the state $\ket{\Psi}$ has definite energy (in the case of a free particle the definite energy states are plane waves) then we can replace $H\ket{\Psi}$ with $E\ket{\Psi}$ and we get $$i \hbar \frac{d\ket{\Psi}}{dt} = E \ket{\Psi} \, . \tag{6}$$ From here, if we assume that the state has a sinusoidal time dependence $\exp(-i \omega t)$ then we get $E = \hbar \omega$.
That time dependence can be assumed because it is a solution to the equation, and any other solution can be written as a linear superposition of such solutions. If you haven't learned about this idea yet don't worry about it. Note, however, that you could also choose $\exp(i \omega t)$ in which case you get $E = -\hbar \omega$. I won't get into the meaning of positive and negative energies in quantum mechanics in this post. If you're interested in that please ask a separate question though, because it is an interesting topic.